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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Hard
  • Numerical

A simple pendulum with length \(100\,cm\) and bob of mass \(250\,g\) is executing S.H.M. of amplitude \(10\,cm\). The maximum tension in the string is found to be \(\frac{x}{40}\,N\). The value of \(x\) is \(..........\).

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Answer & Solution

Correct Answer

(C) \(99\)

Step-by-step Solution

Detailed explanation

\(\sin \theta_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}\)

From conservation of energy

\(\frac{1}{2} m v^2=m g l(1-\cos \theta)\)

Maximum tension occurs at mean position.

\(\therefore T-m g=\frac{m v^2}{l}\)

\(\Rightarrow T=m g+\frac{m v^2}{l}\)

\(\therefore T=m g+2 m g(1-\cos \theta)\)

\(=m g\left[1+2\left(1-\sqrt{1-\sin ^2 \theta}\right)\right]\)

\(=m g\left[3-2 \sqrt{1-\frac{1}{100}}\right]\)

\(=\frac{250}{1000} \times 9.8\left[3-2\left(1-\frac{1}{200}\right)\right]=\frac{99}{40}\)

\(\therefore x=99\)

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